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A295321 Sum of the products of the smaller and larger parts of the partitions of n into two distinct parts with the larger part even. 2

%I #33 Dec 02 2023 22:48:09

%S 0,0,2,0,4,8,18,12,26,40,64,52,82,112,156,136,188,240,310,280,360,440,

%T 542,500,614,728,868,812,966,1120,1304,1232,1432,1632,1866,1776,2028,

%U 2280,2570,2460,2770,3080,3432,3300,3674,4048,4468,4312,4756,5200,5694

%N Sum of the products of the smaller and larger parts of the partitions of n into two distinct parts with the larger part even.

%C Sum of the areas of the distinct rectangles with even length and integer width such that L + W = n, W < L. For example, a(12) = 52; the rectangles are 2 X 10 and 4 X 8 (6 X 6 is not included since we have W < L), so 2*10 + 4*8 = 52.

%C Sum of the ordinates from the ordered pairs (n-k,n*k-k^2) corresponding to integer points along the right side of the parabola b_k = n*k-k^2 where n-k is an even integer such that 0 < k < floor(n/2).

%C Sum of the areas of the trapezoids with bases n and n-2i and height i for even values of n-i where i is in 0 <= i <= floor((n-1)/2). For a(n) the area formula for a trapezoid becomes (n+n-2i)*i/2 = (2n-2i)*i/2 = i*(n-i). For n=10, n-i is even when i=0,2,4 so a(10) = 0*(10-0) + 2*(10-2) + 4*(10-4) = 0 + 16 + 24 = 40. - _Wesley Ivan Hurt_, Mar 22 2018

%C Sum of the areas of the symmetric L-shaped polygons with long side n/2 and width i such that n-i is even for i in 0 <= i <= floor((n-1)/2). The area of each polygon is given by i^2+2i(n/2-i) = i^2+ni-2i^2 = i(n-i). For n=9, 9-i is even for i=1,3 so 1(9-1) + 3(9-3) = 8 + 18 = 26. - _Wesley Ivan Hurt_, Mar 26 2018

%H Vincenzo Librandi, <a href="/A295321/b295321.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%F a(n) = Sum_{i=1..floor((n-1)/2)} i * (n-i) * ((n-i+1) mod 2).

%F Conjectures from _Colin Barker_, Nov 20 2017: (Start)

%F G.f.: 2*x^3*(1 - x + 2*x^2 + 2*x^3 + 2*x^4 + x^6 + x^7) / ((1 - x)^4*(1 + x)^3*(1 + x^2)^3).

%F a(n) = a(n-1) + 3*a(n-4) - 3*a(n-5) - 3*a(n-8) + 3*a(n-9) + a(n-12) - a(n-13) for n>13.

%F (End)

%F a(n) = (n*(4*n^2-3*n-4)/3-n*(n+4)*(-1)^n-(2*n^2+1)*(-1)^floor(n/2)+(-1)^floor((n+1)/2))/32. - _Wesley Ivan Hurt_, Dec 02 2023

%e a(16) = 136; the partitions of 16 into two distinct parts are (15,1), (14,2), (13,3), (12,4), (11,5), (10,6), (9,7). There are 3 partitions with the larger part even, and the sum of the products of the smaller and larger parts is then 14*2 + 12*4 + 10*6 = 136.

%p A295321:=n->add(i*(n-i)*((n-i+1) mod 2), i=1..floor((n-1)/2)): seq(A295321(n), n=1..100);

%t Table[Sum[i (n - i) Mod[n - i + 1, 2], {i, Floor[(n - 1)/2]}], {n, 80}]

%o (PARI) a(n) = sum(i=1, (n-1)\2, i*(n-i)*((n-i+1) % 2)); \\ _Michel Marcus_, Mar 26 2018

%Y Cf. A295320.

%K nonn,easy

%O 1,3

%A _Wesley Ivan Hurt_, Nov 19 2017

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